Hybrid Discrete Teaching-learning-based Optimization Algorithm for Solving Complex Parallel Machine Scheduling Problem
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摘要: 针对制造行业中广泛存在的一类复杂并行机调度问题, 即带到达时间、多工序、加工约束和序相关设置时间的并行机调度问题(Parallel machine scheduling problem with arrival time, multiple operations, process restraints and sequence-dependent setup times, PMSP_AMPS), 建立问题的排序模型并提出一种混合离散教与学优化算法进行求解, 优化目标为最小化最大完工时间.首先, 根据标准教与学算法(Teaching-learning-based optimization, TLBO)中两阶段个体更新公式的特点, 在保留每一阶段个体更新公式框架不变的前提下, 对公式中具体改变实数个体或向量的每个核心操作均用所设计的排列操作进行替换, 使其可直接在离散问题解空间中执行基于标准教与学算法机理的全局搜索, 从而明显提高了原算法的全局搜索效率.其次, 采用交换操作和插入操作构造了一种简洁有效地变邻域局部搜索, 对全局搜索发现的优质解区域进行细致搜索, 从而进一步增强了算法的性能.通过对不同测试问题的仿真实验和算法比较, 验证了所提算法可有效求解PMSP_AMPS.Abstract: This paper proposes a hybrid discrete teaching-learning-based optimization algorithm and set up the scheduling model for solving the parallel machine scheduling problem with arrival time, multiple operations, process restraints, and sequence-dependent setup times (PMSP_AMPS), which widely exists in the manufacturing process. The criterion is to minimize the maximum completion time. Firstly, according to the characteristics of the two stage individual updating formulas in the standard teaching-learning-based optimization (TLBO) algorithm, under the premise of keeping the framework of individual updating formulas at each stage unchanged, each core operation that specifically changes the real number or vector in the formula is replaced with proposed arrangement operation, so that it can execute the global search directly in the discrete solution space based on the TLBO mechanism, which significantly improve the TLBO's global exploration ability. Secondly, a simple but effective variable-neighborhood local search is constructed by using interchange and insert operations, which can perform a detailed search of the high-quality solution regions found by the global search, thereby further enhancing the performance of the algorithm. Simulation experiments and comparisons on different instances demonstrate that the proposed algorithm can effectively solve PMSP_AMPS.
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Key words:
- Parallel machine scheduling problem (PMSP) /
- multiple operations /
- sequence-dependent setup times /
- arrival times /
- discrete teaching-learning-based optimization
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图的同构判定在有机化学、计算机科学和人工智能(机器学习或模式识别中常使用图的邻接矩阵特征值来判定两个对象是否为同一个对象)等领域有着重要的应用[1−3]. 上世纪前半叶, 与图同构相关的问题主要围绕图的邻接矩阵特征值及其应用展开[4], 取得若干重要的理论成果和一批应用成果; 其中一个重要的发现是, 两个互不同构的无向图却具有相同的邻接矩阵特征多项式, 即用邻接矩阵特征多项式判定两个无向图是否同构有时会得出错误的结果[4].
上世纪70年代, Karp和Johnson认为图的同构判定问题是少数几个既不能归类为 P 问题也不能归类为 NP 的问题[1, 3]. 此后, 该问题成为理论计算机领域的公开问题并受到广泛重视. 1982年, Luks给出一个当时最好的两图同构判定算法, 其时间复杂度是$\exp ({\text{O(}}\sqrt {n\log n} {\text{)}}$)[5−6] ($ n $为图的顶点数). 此后40多年来, 几百篇这方面的文章得以在不同的学术期刊上发表[1]. 2015年, Babai[6]在Luks算法的基础上, 给出两图同构判定问题的拟多项式算法, 其时间复杂度是$ \exp ({(\log n)^{{\text{O(1)}}}}) $. Babai的工作虽然是一个重要进展, 但图的同构判定问题是否在多项式时间内可解仍然悬而未决[1, 3].
图的同构判定问题的另一研究路径是设计可执行的判定算法, 大致可以分为传统和非传统两类判定算法. 传统判定算法有两类: 1) 针对一些特殊图[7−8] (如树和极大外平面图)的同构判定算法(如Ullman算法和Schmidt算法等[9]). 2) 针对一般简单图的同构判定算法, 如一些放在因特网上用于测试两图是否同构的程序(如Nauty和Saucy等[3, 6]). 非传统判定算法也有两类: 1) 基于遗传算法和神经网络的智能判定算法. 这些算法将图的同构判定问题转化为一类优化求解问题, 但其判定结果并不完全可靠. 2) 基于DNA计算[9]和量子计算的判定算法. 这类算法虽然高效, 但不能回答图的同构判定是 P 问题还是 NP 问题. 无向连通图的距离矩阵是素矩阵(邻接矩阵一般不是素矩阵), 它具有邻接矩阵所没有的独特性质[10]. 文献[10]揭示了简单无向图同构与距离矩阵同构之间的一般关系, 将基于邻接矩阵的同构判定条件推广到距离矩阵, 给出几个适合一般简单无向图的同构判定条件. 这些条件均是充要条件且均具有多项式时间复杂度. 上述成果是近年来图同构研究方面的一个重要进展.
简单无向图的同构关系可由其距离矩阵的同构关系确定, 但复杂无向图却不能. 其根本原因在于: 复杂无向图中的自环和顶点间的重边对顶点间的距离没有影响[10]; 换言之, 自环数或重边数不同的复杂无向图可能具有完全相同的距离矩阵. 因文献[10]所给的判定条件不能用于复杂无向图, 故寻找和发现复杂无向图的同构判定条件既是必须的也是重要的. 本文的主要贡献是: 1) 针对一般复杂无向图的同构判定问题, 给出了基于邻接矩阵之和的特征多项式判定条件. 2) 针对复杂无向连通图的同构判定问题, 给出了基于距离矩阵特征多项式和邻接矩阵特征多项式的同构判定条件. 将该条件用于复杂无向不连通图的各个连通子图, 就可解决复杂无向不连通图的同构判定问题. 上述两个判定条件均是充要条件且均具有多项式时间复杂度. 此外, 当复杂无向图退化为简单无向图时, 上述两个判定条件仍然适用.
1. 相关概念和预备知识
定义1[4, 11−12]. 图$ G $是一个二元组, 记作$ G = \langle V, E \rangle $, 其中:
1) $ V = \{ {v_1},\;{v_2},\; \cdots ,\;{v_n}\left| {\left| V \right| = n,\;{\text{ 1}} \le n < \infty } \right.\} $, $ {v_i} \in V $称为$ G $的顶点, $ V $称为$ G $的顶点集.
2) $ E = \{ {e_1},\;{e_2},\; \cdots ,\;{e_m}\left| {\left| E \right| = m,\;{\text{ 0}} \le m < \infty } \right.\} $, $ {e_j} \in E $称为$ G $的边, $ E $称为$ G $的边集.
3) $ \forall {e_j} \in E $: $ {e_j} $为无向边时, 称$ G $为无向图, $ {e_j} $为有向边时, 称$ G $为有向图.
4) 连接同一顶点的边称为自环, 两顶点间的多条无向边或多条方向相同的有向边称为重边. 既无自环也无重边的图称为简单图, 否则称为复杂图.
在现有文献报道的复杂无向图中, 每个顶点最多有一个自环. 若无特别声明, 本文讨论的均是顶点最多有一个自环的复杂无向图.
定义2[11−13]. 设$ G = \left\langle {V,\;E} \right\rangle $是无向图, $ V = \{ {v_1}, {v_2},\; \cdots ,\;{v_n}\} $. 若$ {v_i} $和$ {v_j} $之间有$ k $ ($ k = 0,\;1,\; \cdots $)条边, 令$ {a_{ij}} = k $; 称由元素$ {a_{ij}} $构成的矩阵$ A({a_{ij}}) \in {{\bf{R}}^{n \times n}} $为无向图$ G $的邻接矩阵.
定义2既适合简单无向图也适合复杂无向图($ {a_{ij}} \in \{ 0,\;1,\;2,\; \cdots \} $). 无向图的邻接矩阵是非负整数对称矩阵.
定义3[14]. 设$ {I_n} \in {{\bf{R}}^{n \times n}} $为单位矩阵, 交换$ {I_n} $的第$ i $行和第$ j $行(或第$ i $列与第$ j $列)所得的矩阵$ P(i,\;j) $称为对换矩阵, 对换矩阵的乘积称为置换矩阵.
引理1[14]. 对换矩阵和置换矩阵具有如下性质:
1) 置换矩阵的乘积仍是置换矩阵;
2) 设$ Q $是置换矩阵, 则$ \det (Q) = \pm 1 $;
3) 设$ Q $是置换矩阵, 则$ {Q^{\text{T}}} $和$ {Q^{ - 1}} $也是置换矩阵, 且$ {Q^{\text{T}}} = {Q^{ - 1}} $;
4) 置换矩阵是正交矩阵;
5) 置换矩阵是幂幺矩阵, 即若$ Q $是置换矩阵, 则$ {Q^m} = {I_n} $, $ m $是某自然数.
由定义3和引理1可知, 对换矩阵和$ {I_n} $也是置换矩阵.
定义4[10]. 设$ A,\;B \in {{\bf{R}}^{n \times n}} $, $ \Omega \subset {{\bf{R}}^{n \times n}} $为全体$ n $阶置换矩阵的集合. 若存在置换矩阵$ Q \in \Omega $, 使得$ A = {Q^{\text{T}}}BQ $, 则称$ A $与$ B $同构, 记作$ A \cong B $; 此外, 称$ {Q^{\text{T}}}BQ $是对$ B $的同构变换.
图的同构问题可以这样表述: 给定两个图, 当忽略图中顶点标号、顶点间相对位置、边的长短和曲直信息时, 问这两个图是否具有相同的结构?
定义5[4, 12−13]. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_2} = $ $ \langle {V_2}, {E_{\text{2}}} \rangle $是两个图. 若存在一个从$ {V_1} $到$ {V_2} $的一一映射$ g $: $ \forall {v_i},\;{v_j} \in {V_1} $, $ {v_i} $至$ {v_j} $有$ k $条边, 当且仅当$ g({v_i}), g({v_j}) \in {V_2} $, 且$ g({v_i}) $至$ g({v_j}) $也有$ k $条边, 则称$ {G_1} $和$ {G_2} $同构, 记作$ {G_1} \cong {G_2} $.
定义5既适合简单图也适合复杂图($ k \in \{ 0, 1,\;2,\; \cdots \} $).
引理2[10−11]. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_2} = $$ \left\langle {{V_{\text{2}}},\;{E_{\text{2}}}} \right\rangle $是两个图, $ \left| {{V_1}} \right| = \left| {{V_2}} \right| = n $, $ A $与$ B $分别是$ {G_1} $和$ {G_2} $的邻接矩阵, 则$ {G_1} \cong {G_2} $的充要条件是存在$ n $阶置换矩阵$ Q \in \Omega $, 使得$ A = {Q^{\text{T}}}BQ $.
引理2适合一切无向和有向图. 由定义4和引理2可知, 两图同构与该两图的邻接矩阵同构等价.
定义6[4, 10−13]. 设$ G = \left\langle {V,\;E} \right\rangle $是无向图, $ V = \{ {v_1},\;{v_2},\; \cdots ,\;{v_n}\} $, $ E = \{ {e_1},\;{e_2},\; \cdots ,\;{e_m}\} $: 1) $ G $中顶点与边的交替序列$ W = {v_1}{e_1}{v_2}{e_2} \cdots {v_k}{e_k}{v_{k + 1}} $ ($ k \le n - 1 $)称为$ G $的链, 各边互异的链称为迹, 各顶点互异的链称为路, 当$ W $是路时, $ W $中的边数$ k $称为$ W $的长度, 记作$ k = \left| W \right| $; 2) 设$ {v_i} $和$ {v_j} $是$ V $中任意两个顶点, 当$ {v_i} $和$ {v_j} $之间有$ s $条路$ {W_s}{\text{ }} $时, 称$ d({v_i},\;{v_j}) = $$ \min \{ {k_p} = \left| {{W_p}} \right|\left| {1 \le p \le s} \right.\} $为$ {v_i} $和$ {v_j} $之间的距离; 若$ {v_i} $和$ {v_j} $之间无路($ G $不连通), 约定$ d({v_i},\;{v_j}) = \infty $.
定义7[10, 13]. 设$ G = \left\langle {V,\;E} \right\rangle $是无向图, $ V = \{ {v_1}, {v_2},\; \cdots ,\;{v_n}\} $, $ {d_{ij}} = d({v_i},\;{v_j}) $表示顶点$ {v_i} $和$ {v_j} $($ 1 \le i, j \le n $)之间的距离: 当$ i = j $时, 令$ {d_{ii}} = 0 $; 当$ i \ne j $且$ d({v_i},\;{v_j}) = k $ ($ d({v_i},\;{v_j}) = \infty $)时, 令$ {d_{ij}} = k $$ ({d_{ij}} = \infty ) $, 其中$ k \ge 1 $为正整数, 称由元素$ {d_{ij}} $构成的矩阵$ D({d_{ij}}) \in {{\bf{R}}^{n \times n}} $为无向图$ G $的距离矩阵.
由定义6和定义7可知: 1) 复杂无向图中的自环和顶点间的重边对顶点间的距离没有影响; 2) 当$ G $是无向连通图时, 距离矩阵$ D $是对角线元素均为零而其他元素均为正整数的对称矩阵.
定义8[12−13]. 设$ {G_n} = \left\langle {V,\;E} \right\rangle $是简单无向图, $ V = \{ {v_1},\;{v_2},\; \cdots ,\;{v_n}\} $, $ n \ge 2 $, 若$ \forall i \ne j $, $ {v_i} $和$ {v_j} $之间恰有一条边, 则称$ {G_n} $为无向完全图, 其邻接矩阵记为$ {A_n} $.
由定义2和定义8可知, $ {A_n} $的对角线元素均为0而其他元素均为1. 设$ {G_n} $的距离矩阵为$ {D_n} $, 由定义7可知, $ {D_n} = {A_n} $, 即$ {A_n} $既是$ {G_n} $的邻接矩阵又是$ {G_n} $的距离矩阵.
定义9. 设$ { T} \subset {{\bf{R}}^{n \times n}} $$ (n \ge 2) $表示全体$ n $阶正交矩阵的集合; $ {\Omega ^ - } = \{ - S\left| {S \in \Omega } \right.\} $表示全体$ n $阶负置换矩阵的集合; $ \Phi = \Omega \cup {\Omega ^ - } $表示全体$ n $阶置换矩阵和全体$ n $阶负置换矩阵的并集; $ \Theta = { T} - \Phi $表示$ { T} $中除$ \Phi $之外全体$ n $阶正交矩阵的集合.
由定义9和正交矩阵的性质可知: $ \Theta \cap \Phi = \emptyset $($ \emptyset $为空集), $ \Theta \cup \Phi = { T} $; $ \forall {Q_1},\;{Q_2} \in \Phi $, $ {Q_1}{Q_2} \in \Phi $; $ \forall Q \in \Phi $, $ \forall P \in \Theta $, $ QP,\;{\text{ }}PQ \in \Theta $; $ \forall P \in \Theta $, $ \forall E \in \Phi $, $ \forall Q \in { T} $且$ Q \ne E{P^{\text{T}}} $, $ QP,\;{\text{ }}PQ \in \Theta $.
2. 复杂无向图的同构判定方法
为简便计, $ \forall A \in {{\bf{R}}^{n \times n}} $$ (n \ge 2) $, 本文用$ {f_\lambda }(A) $表示$ A $的特征多项式 $ \det (\lambda {I_n} - A) $, 即$ {f_\lambda }(A) = \det (\lambda {I_n} - A) $.
推论1. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_{\text{2}}} = \left\langle {{V_{\text{2}}},\;{E_{\text{2}}}} \right\rangle $是两个无向图, $ \left| {{V_1}} \right| = \left| {{V_2}} \right| = n $, $ A $与$ B $分别是$ {G_1} $和$ {G_2} $的邻接矩阵, 若$ {f_\lambda }(A) \ne {f_\lambda }(B) $, 则$ {G_1} $和$ {G_2} $不同构.
证明. 设$ {f_\lambda }(A) \ne {f_\lambda }(B) $时, $ {G_1} \cong {G_2} $. 由引理2可知, 当$ {G_1} \cong {G_2} $时, 存在$ n $阶置换矩阵$ Q \in \Omega $, 使得$ A = {Q^{\text{T}}}BQ $. 由此, $ {f_\lambda }(A) = $ $ {f_\lambda }({Q^{\text{T}}}BQ) = {f_\lambda }(B) $, 与假设相矛盾. 故$ {f_\lambda }(A) \ne {f_\lambda }(B) $时, $ {G_1} $和$ {G_2} $不同构.
□ 推论1既适合简单无向图也适合复杂无向图. 需要强调的是, 即使$ {f_\lambda }(A) = {f_\lambda }(B) $, 也不能保证$ {G_1} $和$ {G_2} $同构. 邻接矩阵特征多项式相等仅是两个无向图同构的必要条件[10, 15].
设$ A({a_{ij}}) \in {{\bf{R}}^{n \times n}} $$ (n \ge 2) $是对称矩阵: 若$ {a_{ij}} $全是正整数, 则称$ A $为正整数对称矩阵; 若$ {a_{ij}} $为零或正整数, 则称$ A $为非负整数对称矩阵.
引理3[10]. 设$ A \in {{\bf{R}}^{n \times n}} $$ (n \ge 2) $是对角线元素均为1而其他元素均为正整数的对称矩阵, 则$ \forall P \in \Theta $, $ {P^{\text{T}}}AP $不是正整数对称矩阵.
引理4[14]. 设$ A \in {{\bf{R}}^{n \times n}} $, 则$ A $为正交矩阵的充要条件是存在正交矩阵$ P \in { T} $, 使得
$$ {P^{\text{T}}}AP = {P^{ - 1}}AP = \left[ {\begin{array}{*{20}{c}} {{I_s}}&{}&{} \\ {}&{ - {I_t}}&{} \\ {}&{}&{{H_\theta }} \end{array}} \right] = {P_\theta } $$ 其中, $ {I_s} $与$ I_{t} $分别是$ s $阶和$ t $阶单位矩阵,
$$ {H_\theta } = \left[ {\begin{array}{*{20}{c}} {\left[ {\begin{array}{*{20}{c}} {\cos {\theta _1}}&{\sin {\theta _1}} \\ { - \sin {\theta _1}}&{\cos {\theta _1}} \end{array}} \right]}&{}&{} \\ {}& \ddots &{} \\ {}&{}&{\left[ {\begin{array}{*{20}{c}} {\cos {\theta _k}}&{\sin {\theta _k}} \\ { - \sin {\theta _k}}&{\cos {\theta _k}} \end{array}} \right]} \end{array}} \right] $$ $ s + t + 2k = n $, $ {\theta _j} \in {\bf{R}} $$ (1 \le j \le k) $为实数.
由正交矩阵的定义可知, $ {P_\theta } $和$ {H_\theta } $均是正交矩阵.
定理1. 设$ A \in {{\bf{R}}^{n \times n}} $$ (n \ge 2) $是对角线元素为0或1而其他元素均为正整数的对称矩阵, 则$ \forall P \in \Theta $, $ {P^{\text{T}}}AP $既不是正整数对称矩阵也不是非负整数对称矩阵.
证明. 当$ A $的对角线元素全为1时, 由引理3可知, $ \forall P \in \Theta $, $ {P^{\text{T}}}AP $不是正整数对称矩阵. 当$ A $的对角线元素全为0时, 设$ B = A + {I_n} $. 由引理3可知, $ \forall P \in \Theta $, $ {P^{\text{T}}}BP $不是正整数对称矩阵. 因$ {P^{\text{T}}}BP = {P^{\text{T}}}AP + {I_n} $, 故$ \forall P \in \Theta $, $ {P^{\text{T}}}AP $不是非负整数对称矩阵. 当$ A $的对角线元素不全为0也不全为1时, 以下用归纳法证明: $ \forall P \in \Theta $, $ {P^{\text{T}}}AP $不是非负整数对称矩阵.
当$ n = 2 $时, 设
$$ A = \left[ {\begin{array}{*{20}{c}} {{\delta _1}}&a \\ a&{{\delta _2}} \end{array}} \right] ,\; P = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta } \\ { - \sin \theta }&{\cos \theta } \end{array}} \right] $$ 其中, $ {\delta _1},\;{\delta _2} \in \{ 0,\;1\} $, $ {\delta _1} + {\delta _2} = 1 $, $ a \ge 1 $为正整数, $ \theta \in {\bf{R}} $. 设$ {\vartheta _1} = \sin 2\theta $, $ {\vartheta _2} = \cos 2\theta $, $ C = {P^{\text{T}}}AP $, 可得
$$ C = \left[ {\begin{array}{*{20}{c}} {{{\sin }^2}\theta + {\delta _1}{\vartheta _2} - a{\vartheta _1}}&{({\delta _1} - 0.5){\vartheta _1} + a{\vartheta _2}} \\ {({\delta _1} - 0.5){\vartheta _1} + a{\vartheta _2}}&{{{\cos }^2}\theta - {\delta _1}{\vartheta _2} + a{\vartheta _1}} \end{array}} \right] $$ 当$ \theta \ne \pm k\pi $$ (k = 0,\;1,\;2,\; \cdots ) $时, $ P \in \Theta $, $ C $不是非负整数对称矩阵($ C $的元素中有负数或小数). 当$ P \in \Theta $时, 由定义9和引理4可知, 对一切2阶正交矩阵$ E \in \Phi $, $ Z \in {\rm T} $且$ Z \ne E{P^{ - 1}} $, 则: 1) $ {J_1} = ZP \in \Theta $, $ {J_2} = PZ \in \Theta $, $ {J_3} = {Z^{\text{T}}}PZ \in \Theta $; 2) $ J_1^{\text{T}}A{J_1} $, $ J_2^{\text{T}}A{J_2} $, $ J_3^{\text{T}}A{J_3} $均不是非负整数对称矩阵. 由此, 对一切2阶正交矩阵$ J \in \Theta $, $ {J^{\text{T}}}AJ $不是非负整数对称矩阵.
当$ n = 3 $时, 设
$$ A = \left[ {\begin{array}{*{20}{c}} {{\delta _1}}&{{a_1}}&{{a_2}} \\ {{a_1}}&{{\delta _2}}&{{a_3}} \\ {{a_2}}&{{a_3}}&{{\delta _3}} \end{array}} \right] ,\; P = \left[ {\begin{array}{*{20}{c}} \delta &0&0 \\ 0&{\cos \theta }&{\sin \theta } \\ 0&{ - \sin \theta }&{\cos \theta } \end{array}} \right] $$ 其中, $ {\delta _1},\;{\delta _2},\;{\delta _3} \in \{ 0,\;1\} $, $ 1 \le {\delta _1} + {\delta _2} + {\delta _3} \le 2 $, $ {a_i} \ge 1 $ $ (1 \le i \le 3) $为正整数, $ \delta = \pm 1 $, $ \theta \in {\bf{R}} $. 设$ {\vartheta _1} = \sin 2\theta $, $ {\vartheta _2} = \cos 2\theta $, $ \alpha (\theta ) = {a_1}\cos \theta - {a_2}\sin \theta $, $ \beta (\theta ) = {a_1}\sin \theta $ $ +{a_2}\cos \theta $, $ \eta = {\delta _2} - {\delta _3} $, $ \mu (\theta ) = \eta {\sin ^2}\theta $, $ C = {P^{\mathrm{T}}}AP $, 可得
$$ C = \left[ {\begin{array}{*{20}{c}} {{\delta _1}}&{\delta \alpha (\theta )}&{\delta \beta (\theta )} \\ {\delta \alpha (\theta )}&{{\delta _2} - \mu (\theta ) - {a_3}{\vartheta _1}}&{0.5\eta {\vartheta _1} + {a_3}{\vartheta _2}} \\ {\delta \beta (\theta )}&{0.5\eta {\vartheta _1} + {a_3}{\vartheta _2}}&{{\delta _3} + \mu (\theta ) + {a_3}{\vartheta _1}} \end{array}} \right] $$ 当$ \delta = 1 $且 $ \theta \ne \pm 2l\pi $$ (l = 0,\;1,\;2,\; \cdots ) $, 或者$ \delta = - 1 $且$ \theta \ne \pm h\pi $ ($ h $为奇数)时, $ P \in \Theta $, $ C $不是非负整数对称矩阵. 类似$ n = 2 $时的分析可得, 对一切3阶正交矩阵$ J \in \Theta $, $ {J^{\text{T}}}AJ $不是非负整数对称矩阵.
当$ n \ge 4 $时, 设
$$ A = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}&{{A_{13}}} \\ {A_{12}^{\text{T}}}&{{A_{22}}}&{{A_{23}}} \\ {A_{13}^{\text{T}}}&{A_{23}^{\text{T}}}&{{A_{33}}} \end{array}} \right] ,\; {P_\theta } = \left[ {\begin{array}{*{20}{c}} {{I_s}}&{}&{} \\ {}&{ - {I_t}}&{} \\ {}&{}&{{H_\theta }} \end{array}} \right] $$ 其中, $ {A_{11}} \in {{\bf{R}}^{s \times s}} $、$ {A_{22}} \in {{\bf{R}}^{t \times t}} $和$ {A_{33}} \in {{\bf{R}}^{2k \times 2k}} $均是对角线元素为0或1而其他元素全是正整数的分块对称矩阵; $ {A_{12}} $, $ {A_{13}} $, $ {A_{23}} $均是分块正整数矩阵; $ A $的对角线元素之和满足$ 1 \le \sum\nolimits_i {{a_{ii}}} \le n - 1 $; $ {H_\theta } $为形如引理4中的矩阵, $ s + t + 2k = n $, $ {\theta _j} \in {\bf{R}} $$ (1 \le j \le k) $为实数. 由此,
$$ P_\theta ^{\text{T}}A{P_\theta } = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{ - {A_{12}}}&{{A_{13}}{H_\theta }} \\ { - A_{12}^{\text{T}}}&{{A_{22}}}&{ - {A_{23}}{H_\theta }} \\ {H_\theta ^{\text{T}}A_{13}^{\text{T}}}&{ - H_\theta ^{\text{T}}A_{23}^{\text{T}}}&{H_\theta ^{\text{T}}{A_{33}}{H_\theta }} \end{array}} \right] $$ 1) 当$ s = n $, 或$ 1 \le s < n $, $ t = 0 $且所有的$ {\theta _j} = \pm 2l\pi $$ (l = 0,\;1,\;2,\; \cdots ) $, 或$ s = t = 0 $且所有的$ {\theta _j} = \pm 2l\pi $$ (l = 0,\;1,\;2,\; \cdots ) $时, $ {P_\theta } = {I_n} \in \Omega $.
2) 当$ t = n $, 或$ 1 \le t < n $, $ s = 0 $且所有的$ {\theta _j} = \pm h\pi $($ h $为奇数), 或$ t = s = 0 $且所有的$ {\theta _j} = \pm h\pi $($ h $为奇数)时, $ {P_\theta } = - {I_n} \in \Phi $.
3) 当$ s,\;t,\;k $和$ {\theta _j} $$ (1 \le j \le k) $的取值不是上述两种情况时, $ {P_\theta } \in \Theta $, $ P_\theta ^{\text{T}}A{P_\theta } $不是非负整数对称矩阵.
当$ P \in \Theta $时, 由定义9和引理4可知, 对一切$ n \ge 4 $阶正交矩阵$ E \in \Phi $, $ Z \in {T} $且$ Z \ne E{P^{ - 1}} $, 则: a) $ {J_1} = ZP \in \Theta $, $ {J_2} = PZ \in \Theta $, $ {J_3} = {Z^{\text{T}}}PZ \in \Theta $; b) $ J_1^{\text{T}}A{J_1} $, $ J_2^{\text{T}}A{J_2} $, $ J_3^{\text{T}}A{J_3} $均不是非负整数对称矩阵. 由此, 对一切$ n \ge 4 $阶正交矩阵$ J \in \Theta $, $ {J^{\text{T}}}AJ $不是非负整数对称矩阵.
综上可得, $ \forall n \ge 2 $, $ \forall P \in \Theta $, $ {P^{\text{T}}}AP $不是非负整数对称矩阵.
□ 此外, 不难证明, $ \forall n \ge 2 $, $ \forall Q \in \Phi $, $ {Q^{\text{T}}}AQ $是非负整数对称矩阵.
使用定义8中的符号$ {A_n} $和引理2, 可以得到如下推论.
推论2. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_2} = \left\langle {{V_{\text{2}}},\;{E_{\text{2}}}} \right\rangle $是两个无向图, $ \left| {{V_1}} \right| = \left| {{V_2}} \right| = n \ge 2 $, $ A $与$ B $分别是$ {G_1} $和$ {G_2} $的邻接矩阵, 则$ {G_1} \cong {G_2} $的充要条件是存在$ n $阶置换矩阵$ Q \in \Omega $, 使得$ A + {A_n} = {Q^{\text{T}}}(B + {A_n})Q $.
证明. 设$ A + {A_n} = {Q^{\text{T}}}(B + {A_n})Q $. 由定义3和引理1可知, 置换矩阵$ Q \in \Omega $$ (Q \ne {I_n}) $可表示为$ m $ ($ m \ge 2 $为正整数)个对换矩阵$ P({i_s},\;{j_s}) = {P_s} $的乘积, 即$ Q = P({i_1},\;{j_1}) \cdots P({i_m},\;{j_m}) = {P_1} \cdots {P_m}$. 此外, 不难验证, 任给对换矩阵$ {P_s} $, $ P_s^{\text{T}}{A_n}{P_s} = {A_n} $. 由此可得: $ {Q^{\text{T}}}{A_n}Q = P_m^{\text{T}} \cdots P_1^{\text{T}}{A_n}{P_1} \cdots {P_m} = {A_n} $; $ A\; + {A_n} = $$ {Q^{\text{T}}}(B + {A_n})Q = {Q^{\text{T}}}BQ + {A_n} $; $ A = {Q^{\text{T}}}BQ $. 再由引理2可得, $ {G_1} \cong {G_2} $.
设$ {G_1} \cong {G_2} $. 由引理2可知, 存在$ n $阶置换矩阵$ Q \in \Omega $, 使得$ A = {Q^{\text{T}}}BQ $. 由上述充分性证明可知, $ \forall Q \in \Omega $, $ {A_n} = {Q^{\text{T}}}{A_n}Q $. 由此, $ A + {A_n} = {Q^{\text{T}}}(B + {A_n})Q $.
□ 容易理解, 推论2适合一切无向图.
引理5[14]. 设$ C,\;B \in {{\bf{R}}^{n \times n}} $是对称矩阵, 则$ {f_\lambda }(C) = {f_\lambda }(B) $的充要条件是存在$ n $阶正交矩阵$ P \in {T} $, 使得$ C = {P^{\text{T}}}BP $.
基于前面的分析和准备, 下面给出两个无向图的同构判定条件.
定理2. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_2} = \left\langle {{V_{\text{2}}},\;{E_{\text{2}}}} \right\rangle $是两个无向图, 每个顶点最多有一个自环, $ \left| {{V_1}} \right| = \left| {{V_2}} \right| = n \ge 2 $, $ {A_1} $与$ {A_2} $分别是$ {G_1} $和$ {G_2} $的邻接矩阵, 则$ {G_1} \cong {G_2} $的充要条件是$ {f_\lambda }({A_1} + {A_n}) = {f_\lambda }({A_2} + {A_n}) $.
证明. 设$ {f_\lambda }({A_1} + {A_n}) = {f_\lambda }({A_2} + {A_n}) $. 给$ {G_1} $和$ {G_2} $中的每两个顶点之间加一条边, 可以得到两个新的无向图$ {\hat G_1} $和$ {\hat G_2} $, 其邻接矩阵分别是$ {A_1} + {A_n} $和$ {A_2} + {A_n} $. 因$ {G_1} $和$ {G_2} $的顶点最多有一个自环, 由定义2可知, $ {A_1} $和$ {A_2} $的对角线元素为0或1 (可以全为0, 也可全为1); 再由定义8可知, $ {A_1} + {A_n} $和$ {A_2} + {A_n} $均是对角线元素为0或1而其他元素均是正整数的对称矩阵, 即$ {A_1} + {A_n} $和$ {A_2} + {A_n} $均是非负整数对称矩阵. 由引理5可知, 当$ {f_\lambda }({A_1} + {A_n}) = {f_\lambda }({A_2} + {A_n}) $时, 存在正交矩阵$ Q \in { T} $, 使得$ {A_1} \;+ {A_n} = {Q^{\text{T}}}({A_2} + {A_n})Q $. 由定理1可知, $ Q \notin \Theta $; 否则, $ {A_1} + {A_n} $将不是非负整数矩阵. 因$ Q \in { T} = \Phi \cup \Theta $且$ Q \notin \Theta $, 故$ Q \in \Phi $. 由此, 存在$ Q \in \Omega $ (或$ - Q \in \Omega $), 使得$ {A_1} + {A_n} = {Q^{\text{T}}}({A_2} + {A_n})Q $; 再由推论2可知, $ {G_1} \cong {G_2} $.
设$ {G_1} \cong {G_2} $. 因$ {G_1} \cong {G_2} $, 故存在$ n $阶置换矩阵$ Q \in \Omega $, 使得$ {A_1} = {Q^{\text{T}}}{A_2}Q $. 由推论2的证明结果可知, $ {A_n} = {Q^{\text{T}}}{A_n}Q $. 由此, $ {A_1} + {A_n} = $$ {Q^{\text{T}}}({A_2} + {A_n})Q $, $ {f_\lambda }({A_1} + {A_n}) = {f_\lambda }({A_2} + {A_n}) $.
□ 需要说明的是: 1) 定理2既适合复杂无向图也适合简单无向图, 既适合连通无向图也适合不连通无向图; 一言以蔽之, 定理2适合一切无向图. 2) 因求解特征多项式$ {f_\lambda }(A) $的算法复杂度是$ {\mathrm{O}}({n^3}) $[16−17], 故使用定理2判定两个无向图是否同构的算法复杂度是$ 2{\mathrm{O}}({n^3}) $.
引理6[10]. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_{\text{2}}} = \left\langle {{V_{\text{2}}},\;{E_{\text{2}}}} \right\rangle $是两个简单无向连通图, $ \left| {{V_1}} \right| = \left| {{V_2}} \right| = n \ge {\text{2}} $, $ {D_1} $与$ {D_2} $分别是$ {G_1} $和$ {G_2} $的距离矩阵, 则$ {G_1} \cong {G_2} $的充要条件是$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $.
引理7[18]. 设$ C,\;B \in {{\bf{R}}^{n \times n}} $为对称矩阵, 对一切$ 1 \le i \le n $, $ {\lambda _i}(C) $为$ C $的特征值, $ {\lambda _i}(B) $为$ B $的特征值, 则$ \min {\lambda _i}(B) + {\lambda _i}(C) \le {\lambda _i}(C + B) $$ \le {\lambda _i}(C) \;+ \max {\lambda _i}(B) $.
设$ G = \left\langle {V,\;E} \right\rangle $是复杂无向连通图, $ V = \{ {v_1}, {v_2},\; \cdots ,\;{v_n}\} $, $ n \ge 2 $: 1) 除去$ G $中顶点上的自环和邻接顶点间的重边(邻接顶点间仅保留一条边), 可以得到一个简单无向连通图$ {G_1} $(称$ {G_1} $为$ G $的主图). 2) $ \forall i \ne j $, 若$ {v_i} $和$ {v_j} $邻接, 除去$ {v_i} $和$ {v_j} $之间的一条边, 其他保持不变, 可以得到另一个无向图$ {G_2} $(称$ {G_2} $为$ G $的附图). 3) 设$ A $为$ G $的邻接矩阵, $ {A_1} $和$ {A_2} $分别是$ {G_1} $和$ {G_2} $的邻接矩阵, $ D $为$ G $的距离矩阵. 由上述假设可知, $ A = {A_1} + {A_2} $. 因复杂无向图中的自环和顶点间的重边对距离没有影响, 由定义6和定义7可知, $ D $既是$ G $也是$ {G_1} $的距离矩阵.
基于以上的叙述和讨论, 本文给出两个复杂无向图同构的另一个判定条件.
定理3. 设$ {G_{\text{1}}} = \left\langle {{V_{\text{1}}},\;{E_{\text{1}}}} \right\rangle $和$ {G_{\text{2}}} = $$ \left\langle {{V_{\text{2}}},\;{E_{\text{2}}}} \right\rangle $是两个复杂无向连通图, 每个顶点最多有一个自环, $ \left| {{V_1}} \right| = \left| {{V_2}} \right| = n \ge {\text{2}} $, $ {A_1} $与$ {A_2} $分别是$ {G_1} $和$ {G_2} $的邻接矩阵, $ {D_1} $与$ {D_2} $分别是$ {G_1} $和$ {G_2} $的距离矩阵. 则$ {G_1} \cong {G_2} $的充要条件是$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $且$ {f_\lambda }({A_1}) = {f_\lambda }({A_2}) $.
证明. 设$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $且$ {f_\lambda }({A_1}) = $$ {f_\lambda }({A_2}) $. 另设$ {G_{11}} $与$ {G_{12}} $分别是$ {G_1} $的主图和附图, $ {A_{11}} $与$ {A_{12}} $$ ({A_{12}} \ne 0) $分别是$ {G_{11}} $和$ {G_{12}} $的邻接矩阵; $ {G_{21}} $与$ {G_{22}} $分别是$ {G_2} $的主图和附图, $ {A_{21}} $与$ {A_{22}} $$ ({A_{22}} \ne 0) $分别是$ {G_{21}} $和$ {G_{22}} $的邻接矩阵. 由此, 可得: $ {A_1} = {A_{11}} + {A_{12}} $, $ {A_2} = {A_{21}} + {A_{22}} $; $ {D_1} $既是$ {G_1} $也是$ {G_{11}} $的距离矩阵, $ {D_2} $既是$ {G_2} $也是$ {G_{21}} $的距离矩阵. 由引理6可知, 当$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $时, $ {G_{11}} \cong {G_{21}} $; 再由引理2可知, 存在$ Q \in \Omega $, 使得$ {A_{11}} = {Q^{\text{T}}}{A_{21}}Q $. 因$ {A_1} $和$ {A_2} $均是实对称矩阵, 且$ {f_\lambda }({A_1}) = {f_\lambda }({A_2}) $, 由引理5可知, 存在$ n $阶正交矩阵$ P \in { T} $, 使得$ {A_1} = {P^{\text{T}}}{A_2}P $, 即$ {A_{11}} + {A_{12}} = {P^{\text{T}}}{A_{21}}P + {P^{\text{T}}}{A_{22}}P $. 综上可知, 当$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $且$ {f_\lambda }({A_1}) = {f_\lambda }({A_2}) $时
$$ {A_{11}} = {Q^{\text{T}}}{A_{21}}Q $$ (1) 且
$$ {A_{11}} + {A_{12}} = {P^{\text{T}}}{A_{21}}P + {P^{\text{T}}}{A_{22}}P$$ (2) 设$ {A_{12}} = {P^{\text{T}}}{A_{22}}P $, 由式(1)和式(2)可得, $ {A_{11}} = {Q^{\text{T}}}{A_{21}}Q $, $ {A_{11}} = {P^{\text{T}}}{A_{21}}P $, 由此, $ P = Q $. 另由式(2)可得, $ {A_{11}} = {P^{\text{T}}}{A_{21}}P + {P^{\text{T}}}{A_{22}}P - {A_{12}} $.
1) 设$ {H_1} = {P^{\text{T}}}{A_{21}}P $, $ {H_2} = {P^{\text{T}}}{A_{22}}P - {A_{12}} $, 则
$${A_{11}} = {H_1} + {H_2} $$ (3) 因$ {H_1} $和$ {H_2} $均是实对称矩阵且$ \lambda ({H_1}) = \lambda ({A_{21}}) $. 由式(3)和引理7可得
$$ \begin{split} & \min {\lambda _i}({H_2}) + {\lambda _i}({A_{21}}) \le {\lambda _i}({A_{11}}) \le\\& \qquad{\lambda _i}({A_{21}}) +\max {\lambda _i}({H_2}) \end{split} $$ (4) 2) 设$ {A_{12}} \ne {P^{\text{T}}}{A_{22}}P $, 则$ {H_2} \ne 0 $. 因$ {H_2} $是对称矩阵, 故当$ {H_2} \ne 0 $时, $ \lambda ({H_2}) \ne 0 $. 设$ \min {\lambda _i}({H_2}) = \max {\lambda _i}({H_2}) = h $, 由式(4)可得, $ {\lambda _i}({A_{11}}) = {\lambda _i}({A_{21}}) \;+ h $. 若$ \lambda ({A_{11}}) = \lambda ({A_{21}}) $, 则$ h = 0 $, $ \lambda ({H_2}) = 0 $, $ {A_{12}} = {P^{\text{T}}}{A_{22}}P $, 与假设矛盾; 若$ \lambda ({A_{11}}) \ne \lambda ({A_{21}}) $, 则与式(1)矛盾. 设$ {\lambda _i}({A_{11}}) \le $$ {\lambda _i}({A_{21}}) + \max {\lambda _i}({H_2})\; ({\lambda _i}({A_{11}}) \ge {\lambda _i}({A_{21}}) + \min {\lambda _i}({H_2})) $, 若$ \lambda ({A_{11}}) \ne \lambda ({A_{21}}) $, 则与$ {A_{11}} = {Q^{\text{T}}}{A_{21}}Q $矛盾; 若$ \lambda ({A_{11}}) = \lambda ({A_{21}}) $, 则由引理5可得, $ {A_{11}} = {P^{\text{T}}}{A_{21}}P $$ (P \in { T}) $. 由此, ${A_{11}} = {P^{\text{T}}}{A_{21}}P,\; {A_{11}} = {Q^{\text{T}}}{A_{21}}Q $, $ P = Q $. 而当$ P = Q $时, 由式(1)和式(2)可得, $ {A_{12}} = {P^{\text{T}}}{A_{22}}P $, 与$ {A_{12}} \ne {P^{\text{T}}}{A_{22}}P $的假设矛盾. 总之, $ {A_{12}} \ne {P^{\text{T}}}{A_{22}}P $的假设不成立.
综上可知: 当$ {A_{12}} = {P^{\text{T}}}{A_{22}}P $时, $ P = Q $; 由式(2)可得$ {A_1} = {Q^{\text{T}}}{A_2}Q $, 再由引理2可得, $ {G_1} \cong {G_2} $.
设$ {G_1} \cong {G_2} $. 由引理2可知, 存在$ n $阶置换矩阵$ Q \in \Omega $, 使得$ {A_1} = {Q^{\text{T}}}{A_2}Q $. 因$ {G_1} $和$ {G_2} $均是复杂无向连通图, 如充分性证明可设, $ {A_1} = {A_{11}} + {A_{12}} $, $ {A_2} = {A_{21}} + {A_{22}} $, 由此, $ {A_{11}} + {A_{12}} = {Q^{\text{T}}}({A_{21}} + {A_{22}})Q $. 因$ Q $是正交矩阵, 故$ {f_\lambda }({A_1}) = {f_\lambda }({A_2}) $. 设$ {A_{11}} \ne {Q^{\text{T}}}{A_{21}}Q $, 则简单无向连通图$ {G_{11}} $和$ {G_{21}} $不同构, 进而可得$ {G_1} $和$ {G_2} $不同构, 与假设矛盾. 当$ {A_{11}} = {Q^{\text{T}}}{A_{21}}Q $时, 由引理2可知, $ {G_{11}} \cong {G_{21}} $; 再由引理6可得, $ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $. 综上, 当$ {G_1} \cong {G_2} $时, $ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $且$ {f_\lambda }({A_1}) = {f_\lambda }({A_2}) $.
□ 比较引理6和定理3可知, 简单无向连通图的同构条件不同于复杂无向连通图的同构条件.
由定理3的证明过程可知, 当定理3中的$ {G_1} $和$ {G_2} $均是简单无向连通图时, $ {G_1} = {G_{11}} $, $ {G_2} = {G_{21}} $, $ {A_{12}} = {A_{22}} = 0 $, $ {A_1} = {A_{11}} $, $ {A_2} = {A_{21}} $. 由引理6可知: 当$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $时, $ {G_1} \cong {G_2} $; 再由引理2可知, 存在$ Q \in \Omega $, 使得$ {A_1} = {Q^{\text{T}}}{A_2}Q $, 进而可得$ {f_\lambda }({A_1}) = {f_\lambda }({A_2}) $. 由此, 当$ {G_1} $和$ {G_2} $均是简单无向连通图时, 定理3中的条件可简化为$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $; 定理3退化为引理6. 这表明定理3也适合简单无向连通图.
因为求解距离矩阵$ D $的算法复杂度是$ {\mathrm{O}}({n^3}) $[12−13], 求解特征多项式的算法复杂度是$ {\mathrm{O}}({n^3}) $[16−17], 故定理3判定条件的算法复杂度是$ 4{\mathrm{O}}({n^3}) $.
设$ {G_1} $和$ {G_2} $是两个复杂无向不连通图且分别由$ m $个连通子图$ {G_{ij}} $$ (i = 1,\;2;{\text{ }}1 \le j \le m) $组成, $ {G_1} $与$ {G_2} $同构是指$ {G_1} $中的连通子图$ {G_{1j}} $与$ {G_2} $中的连通子图$ {G_{2j}} $一一对应同构. 不难理解, 将定理3用于复杂无向不连通图的各个连通子图, 就可解决复杂无向不连通图的同构判定问题.
比较定理2和定理3可知: 1) 定理2适合一切无向图, 而定理3仅适合复杂无向连通图; 2) 若将定理2限定在复杂无向连通图上, 则定理2与定理3等价; 3) 就一般复杂无向图的同构判定问题而言, 定理2比定理3具有更低的算法复杂度, 因此也更便于实际应用.
1) $ {G_1} $和$ {G_2} $均是复杂无向连通图. 设$ {G_1} $和$ {G_2} $的邻接矩阵分别是$ {A_{11}} $和$ {A_{12}} $, 距离矩阵分别是$ {D_1} $和$ {D_2} $. 按图中顶点标号, 经计算可得
$$ {A_{11}} = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{array}} \right] ,\; {A_{12}} = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&0&2 \\ 1&2&0 \end{array}} \right] $$ $$ {D_1} = {D_2} = \left[ {\begin{array}{*{20}{c}} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{array}} \right] $$ $$ {f_\lambda }({A_{11}} + {A_3}) = {\lambda ^3} - 3{\lambda ^2} - 9\lambda - 5\qquad\qquad $$ $$ {f_\lambda }({A_{12}} + {A_3}) = {\lambda ^3} - {\lambda ^2} - 17\lambda - 15 \qquad\quad\;\;$$ $$ {f_\lambda }({A_{11}}) = {\lambda ^3} - 3{\lambda ^2}, {f_\lambda }({A_{12}}) = {\lambda ^3} - {\lambda ^2} - 6\lambda$$ 其中, $ {A_3} $是3顶点无向完全图的邻接矩阵.
因$ {f_\lambda }({A_{11}} + {A_3}) \ne {f_\lambda }({A_{12}} + {A_3}) $, 由定理2可以判定$ {G_1} $和$ {G_2} $不同构. 因$ {f_\lambda }({D_1}) = {f_\lambda }({D_2}) $, 但$ {f_\lambda }({A_{11}}) \ne {f_\lambda }({A_{12}}) $, 由推论1和定理3均可判定$ {G_1} $和$ {G_2} $不同构.
2) $ {G_3} $和$ {G_4} $均是复杂无向连通图. 设$ {G_3} $的邻接矩阵和距离矩阵分别是$ {A_{31}} $和$ {D_3} $, $ {G_4} $的邻接矩阵和距离矩阵分别是$ {A_{41}} $和$ {D_4} $. 按图中顶点标号, 经计算可得
$$ {A_{31}} = {A_{41}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&0&1&1&1 \\ 0&0&0&0&2&0&1&1 \\ 0&0&0&1&0&1&0&1 \\ 0&0&1&0&1&0&1&0 \\ 0&2&0&1&0&1&0&0 \\ 1&0&1&0&1&0&0&0 \\ 1&1&0&1&0&0&0&0 \\ 1&1&1&0&0&0&0&0 \end{array}} \right] $$ $$ {D_3} = {D_4} = \left[ {\begin{array}{*{20}{c}} 0&2&2&2&2&1&1&1 \\ 2&0&2&2&1&2&1&1 \\ 2&2&0&1&2&1&2&1 \\ 2&2&1&0&1&2&1&2 \\ 2&1&2&1&0&1&2&2 \\ 1&2&1&2&1&0&2&2 \\ 1&1&2&1&2&2&0&2 \\ 1&1&1&2&2&2&2&0 \end{array}} \right] $$ 设$ {A_8} $是8顶点无向完全图的邻接矩阵. 因$ {f_\lambda }({A_{31}} + {A_8}) = {f_\lambda }({A_{41}} + {A_8}) $, 由定理2可以判定$ {G_3} \cong {G_4} $. 因$ {f_\lambda }({D_3}) = {f_\lambda }({D_4}) $且$ {f_\lambda }({A_{31}}) = $$ {f_\lambda }({A_{41}}) $, 故由定理3亦可判定$ {G_3} \cong {G_4} $.
3) $ {G_5} $和$ {G_6} $均是复杂无向连通图. 设$ {G_5} $的邻接矩阵和距离矩阵分别是$ {A_{51}} $和$ {D_5} $, $ {G_6} $的邻接矩阵和距离矩阵分别是$ {A_{61}} $和$ {D_6} $. 按图中顶点标号, 经计算可得
$$ {A_{51}} = {A_{61}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&1&1&1 \\ 0&0&0&1&1&1 \\ 0&0&0&1&1&2 \\ 1&1&1&0&0&0 \\ 1&1&1&0&0&0 \\ 1&1&2&0&0&0 \end{array}} \right] $$ $$ {D_5} = {D_6} = \left[ {\begin{array}{*{20}{c}} 0&2&2&1&1&1 \\ 2&0&2&1&1&1 \\ 2&2&0&1&1&1 \\ 1&1&1&0&2&2 \\ 1&1&1&2&0&2 \\ 1&1&1&2&2&0 \end{array}} \right] $$ 设$ {A_6} $是6顶点无向完全图的邻接矩阵. 因$ {f_\lambda }({A_{51}} + {A_6}) = {f_\lambda }({A_{61}} + {A_6}) $, 由定理2可以判定$ {G_5} \cong {G_6} $. 因$ {f_\lambda }({A_{51}}) = {f_\lambda }({A_{61}}) $且$ {f_\lambda }({D_5}) = $$ {f_\lambda }({D_6}) $, 由定理3亦可判定$ {G_5} \cong {G_6} $.
4) $ {G_7} $和$ {G_8} $均是无向树. 设$ {G_7} $和$ {G_8} $的邻接矩阵分别是$ {A_{71}} $和$ {A_{81}} $, 距离矩阵分别是$ {D_7} $和$ {D_8} $. 按图中顶点标号, 经计算可得
$$ {A_{71}} = \left[ {\begin{array}{*{20}{c}} 0&1&1&1&1&0&0&0 \\ 1&0&0&0&0&1&1&1 \\ 1&0&0&0&0&0&0&0 \\ 1&0&0&0&0&0&0&0 \\ 1&0&0&0&0&0&0&0 \\ 0&1&0&0&0&0&0&0 \\ 0&1&0&0&0&0&0&0 \\ 0&1&0&0&0&0&0&0 \end{array}} \right] \; $$ $$ {A_{81}} = \left[ {\begin{array}{*{20}{c}} 0&1&1&1&1&1&0&0 \\ 1&0&0&0&0&0&0&0 \\ 1&0&0&0&0&0&0&0 \\ 1&0&0&0&0&0&0&0 \\ 1&0&0&0&0&0&0&0 \\ 1&0&0&0&0&0&1&0 \\ 0&0&0&0&0&1&0&1 \\ 0&0&0&0&0&0&1&0 \end{array}} \right] $$ $$ {D_7} = \left[ {\begin{array}{*{20}{c}} 0&1&1&1&1&2&2&2 \\ 1&0&2&2&2&1&1&1 \\ 1&2&0&2&2&3&3&3 \\ 1&2&2&0&2&3&3&3 \\ 1&2&2&2&0&3&3&3 \\ 2&1&3&3&3&0&2&2 \\ 2&1&3&3&3&2&0&2 \\ 2&1&3&3&3&2&2&0 \end{array}} \right] $$ $$ {D_8} = \left[ {\begin{array}{*{20}{c}} 0&1&1&1&1&1&2&3 \\ 1&0&2&2&2&2&3&4 \\ 1&2&0&2&2&2&3&4 \\ 1&2&2&0&2&2&3&4 \\ 1&2&2&2&0&2&3&4 \\ 1&2&2&2&2&0&1&2 \\ 2&3&3&3&3&1&0&1 \\ 3&4&4&4&4&2&1&0 \end{array}} \right] $$ 设$ {A_8} $是8顶点无向完全图的邻接矩阵, 可得$ \det ({A_{71}} + {A_8}) = 17 $, $ \det ({A_{81}} + {A_8}) = 21 $. 由此, $ {f_\lambda }({A_{71}} + {A_8}) \ne {f_\lambda }({A_{81}} + {A_8}) $, 由定理2可以判定$ {G_7} $和$ {G_8} $不同构. 另外, 还可求得: $ {f_\lambda }({A_{71}}) = {f_\lambda }({A_{81}}) = {\lambda ^8} - 7{\lambda ^6} + 9{\lambda ^4} $; $ \det ({D_7}) = $$- 1\,048$, $\det ({D_8}) = 1\,600$. 因$ \det ({D_7}) \ne \det ({D_8}) $, 故$ {f_\lambda }({D_7}) \ne {f_\lambda }({D_8}) $. 由此, $ {f_\lambda }({A_{71}}) = {f_\lambda }({A_{81}}) $, 但$ {f_\lambda }({D_7}) \ne {f_\lambda }({D_8}) $, 由定理3可以判定$ {G_7} $和$ {G_8} $不同构.
3. 结束语
图的同构关系是一种等价关系, 凡与图的结构相关的分类、聚类、识别和学习等问题均与图的同构判定问题有关. 时至今日, 图的同构判定问题仍然具有重要的理论和应用价值.
文献[10]将基于邻接矩阵的同构判定条件(引理2)推广到简单无向图距离矩阵(引理6和引理7), 解决了简单无向图的同构判定问题. 但引理6和引理7不能用于复杂无向图的同构判定, 原因在于复杂无向图的同构关系不能由其距离矩阵唯一确定.
本文的主要思路和贡献可概括为: 1) 证明了对角线元素为0或1而其他元素均为正整数的对称矩阵在非同构正交变换下不再是非负整数对称矩阵(定理1); 利用无向完全图邻接矩阵在矩阵同构变换下保持不变的特性, 给出了引理2的另一种等价表示(推论2); 基于定理1和推论2, 给出了基于邻接矩阵之和的特征多项式判定条件(定理2). 2) 针对复杂无向连通图的同构判定问题, 使用图及邻接矩阵的分解方法, 给出了基于邻接矩阵特征多项式和距离矩阵特征多项式的同构判定条件(定理3). 将该条件用于复杂无向不连通图的各个连通子图, 就可解决复杂无向不连通图的同构判定问题. 定理2和定理3中的判定条件均是充要条件且均具有多项式时间复杂度. 需要再次说明的是: 定理2所给的判定条件具有一般普适性, 即对任意的简单、复杂(每一顶点最多允许有一个自环)、连通和不连通无向图均适用; 定理3既适合复杂无向连通图也适合简单无向连通图.
与文献[10]中的结果相比, 定理2和定理3是无向图同构研究方面的又一重要进展. 定理2和定理3的另一理论价值表明, 无向图的同构判定问题是 P 问题. 需要说明的是, 虽然我们解决了无向图的同构判定问题, 但有向图的同构判定是 P 问题还是 NP 问题仍然没有得到解决.
今后我们将针对有向图的同构判定问题开展研究, 期望得到一些有理论和实用价值的新结果.
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图 1 工序为$ \pi $ = [1 3 2 5 4 1 3 1 3 4]时的甘特图
Fig. 1 The Gantt chart of $ \pi $ = [1 3 2 5 4 1 3 1 3 4]
图 6 各算法在不同时间参数下的均值线及95 %置信度下Tukey$'$s HSD检验的置信区间
Fig. 6 Means plot and 95 % Tukey$'$s HSD confidence intervals for the interaction between the algorithms and the different time factor $ \rho $
图 7 各算法在不同时间参数下的均值线及95 %置信度下Tukey$'$s HSD检验的置信区间
Fig. 7 Means plot and 95 % Tukey$'$s HSD confidence intervals for the interaction between the algorithms and the different time factor $ \rho $
表 1 工序加工约束、序相关设置时间和工件到达时间表($ \pi $ = [1 3 2 5 4 1 3 1 3 4])
Table 1 The schedule of process constraint, sequence setup time and arrival time ($ \pi $ = [1 3 2 5 4 1 3 1 3 4])
可执行操作的设备 序相关设置时间 首次到达设备的时间 操作1 操作2 操作3 工件1 工件2 工件3 工件4 工件5 m1 m2 m3 工件1 m1/m2 m1/m2/m3 m1/m3 — 83 38 39 47 35 34 23 工件2 m1 — — 53 — 66 45 25 38 78 98 工件3 m3 m2/m3 m2 64 57 — 72 46 52 23 32 工件4 m1/m3 m1 — 46 67 83 — 55 132 131 98 工件5 m1/m3 — — 40 66 83 77 — 114 99 112 
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表 2 工件加工时间表($ \pi $ = [1 3 2 5 4 1 3 1 3 4]) ($t$/s)
Table 2 The processing time table ($ \pi $ = [1 3 2 5 4 1 3 1 3 4]) ($t$/s)
操作1 操作2 操作3 m1 m2 m3 m1 m2 m3 m1 m2 m3 工件1 62 44 — 78 86 26 48 — 87 工件2 41 — — — — — — — — 工件3 — — 31 — 58 65 — 42 — 工件4 32 — 31 27 — — — — — 工件5 74 — 36 — — — — — —
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表 3 参数水平设置表
Table 3 Combinations of parameter values
参数 水平 1 2 3 4 $ popsize $ 10 20 30 40 $ TF $ 0 1 2 3 $ r_{m} $ 0.1 0.4 0.7 0.9
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表 4 正交表和AVG统计
Table 4 Orthogonal array and AVG
参数组合 水平 $ AVG $ $ popsize $ $ TF $ $ r_{m} $ 1 1 1 1 500.65 2 1 2 2 503.90 3 1 3 3 497.30 4 1 4 4 498.55 5 2 1 2 489.35 6 2 2 3 491.20 7 2 3 4 485.50 8 2 4 1 488.35 9 3 1 3 482.55 10 3 2 4 481.80 11 3 3 1 482.25 12 3 4 2 483.25 13 4 1 4 483.35 14 4 2 1 484.00 15 4 3 2 483.60 16 4 4 3 482.20
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表 5 各参数响应值
Table 5 Average response value and rank of each parameter
水平 $ popsize $ $ TF $ $ r_{m} $ 1 500.10 488.98 488.81 2 488.60 490.23 490.02 3 482.46 487.16 488.31 4 483.29 488.09 487.30 等级 1 3 2
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表 6 DTLBO与PSO、DTLBO-Ⅰ、标准TLBO和CIWO的比较($ \rho =4 $)
Table 6 Comparisons of DTLBO, DTLBO-Ⅰ, PSO, CIWO, and standard TLBO ($ \rho =4 $)
测试问题 PSO DTLBO-Ⅰ 标准TLBO CIWO DTLBO BST WST AVG BST WST AVG BST WST AVG BST WST AVG BST WST AVG 10$ \times $5 229 235 230.15 228 231 229.45 228 233 230.05 227 246 229.95 227 244 229.3 20$ \times $5 560 587 576.7 546 579 560.3 558 589 574.5 503 554 526.5 506 534 521.4 30$ \times $5 682 708 694.8 638 685 662.3 675 709 697 615 665 635.5 607 638 622.4 40$ \times $5 752 784 771.4 696 748 725.7 728 787 767.75 654 716 690.05 661 716 692.8 50$ \times $5 1 132 1 171 1 157.45 1 093 1 140 1 114.3 1148 1 175 1 161.5 1 019 1 091 1 059.2 1 019 1 078 1 051.85 40$ \times $10 340 357 348.2 307 336 321.3 340 360 351.15 293 326 311.9 284 317 304.3 50$ \times $10 410 429 419.7 373 411 392.75 413 429 421.1 362 391 376.3 353 386 370.75 60$ \times $10 523 548 538.8 497 529 516 538 552 544.3 485 516 499.95 481 516 500.6 70$ \times $10 658 682 672 626 653 641.8 660 684 672.65 604 639 622 597 636 621.75 80$ \times $10 596 619 612.15 567 601 585.55 597 621 612.95 553 589 571.45 553 584 569.5 80$ \times $20 286 298 292.2 265 289 278.85 285 298 291.8 257 282 271.65 259 272 264.7 90$ \times $20 287 299 294.65 271 289 280.55 288 301 297.4 255 284 273.1 255 279 269.05 100$ \times $20 329 340 336 309 328 320.5 333 343 338 304 323 312.65 299 313 306.25 150$ \times $20 479 496 489.2 460 486 472 487 499 492.9 454 476 464.95 449 477 458.45 200$ \times $20 575 593 587.25 557 576 568.5 577 598 589.85 559 575 564.4 547 559 549.55 Average 522.53 543.06 534.71 495.53 525.4 511.32 523.66 545.2 536.19 476.27 511.53 493.97 473.13 503.27 488.84
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表 7 HDTLBO与DPSO、DTLBO-Ⅱ、GA_DR_C和CCIWO的比较($ \rho =4 $)
Table 7 Comparisons of HDTLBO, DPSO, DTLBO-Ⅱ, GA_DR_C, and CCIWO ($ \rho =4 $)
测试问题 DPSO DTLBO-Ⅱ GA_DR_C CCIWO HDTLBO BST WST AVG BST WST AVG BST WST AVG BST WST AVG BST WST AVG 10$ \times $5 227 230 228.75 227 230 228.45 227 230 228.2 227 230 228.2 227 230 227.3 20$ \times $5 521 547 527.15 533 554 543.1 514 551 536.15 494 536 518.1 499 540 520.95 30$ \times $5 612 647 624.1 630 658 645.65 625 656 645.85 617 644 621.95 603 638 617.5 40$ \times $5 652 692 686.75 695 733 710.15 695 735 715.8 657 715 686.15 645 692 681.65 50$ \times $5 1 014 1 074 1 058.4 1 064 1 111 1 090 1 070 1107 1 094.75 1 024 1 088 1 057.05 1 006 1 055 1 035.9 40$ \times $10 303 317 306.14 303 328 316.2 305 325 316.4 292 320 307.7 290 316 303.65 50$ \times $10 365 387 370.95 370 395 387.95 374 400 388.2 362 390 373.85 355 382 369 60$ \times $10 481 503 498.85 498 519 510.9 497 584 541.15 481 515 496.6 481 501 493.05 70$ \times $10 601 639 615.4 626 656 640.15 636 716 694.55 597 639 621.7 597 635 606.75 80$ \times $10 556 590 579 571 596 584.55 561 590 582.75 548 586 568.45 554 575 564.85 80$ \times $20 274 285 277.41 269 285 278 273 285 279.9 275 280 269.3 262 279 273.7 90$ \times $20 269 290 273.33 275 288 282.1 277 292 285.8 268 285 276.25 266 282 272.58 100$ \times $20 304 323 313.7 313 327 320.85 317 328 322.85 300 323 314.6 297 323 310.65 150$ \times $20 472 489 475.34 469 486 476.45 470 488 480.55 469 486 470.7 464 483 469.89 200$ \times $20 553 579 566.5 564 579 573 562 587 578.1 557 583 570.05 549 579 565.2 Average 480.27 506.13 493.45 493.8 516.33 505.83 493.53 524.93 521.73 477.87 508 492.04 473 500.67 487.51
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表 8 模具各工序加工约束及首次到达时间表
Table 8 The schedule of mold process constraint and arrival time
模具 可执行操作的设备 模具首次到达设备时间 操作1 操作2 操作3 m1 m2 m3 m4 m5 1 m1/m3/m5 m3/m4 m2/m3/m4 7 4 8 4 6 2 m1/m3/m6 m1 — 1 6 2 3 3 3 m3 — — 3 1 8 4 6 4 m2/m3/m5 m1/m2/m3/m4/m5 m2/m3/m4 8 1 6 8 7 5 m1/m3/m4 — — 1 2 6 7 3 6 m2/m5 m1/m4/m5 m1/m4/m5 5 8 3 1 5 7 m2/m4/m5 m2/m3/m4/m5 — 3 8 3 3 6 8 m1/m2/m3/m5 — — 2 6 8 7 7 9 m4/m5 m3 m3/m4 6 3 5 3 7 10 m2 m2/m4 m1/m2/m3 7 4 1 4 6 11 m1/m5 m1 — 6 5 1 8 4 12 m4/m5 m1/m2/m3/m4/m5 m1/m2/m4 5 3 1 8 8 13 m2/m4/m5 m3 m4/m5 7 1 6 6 6 14 m3 m1/m3/m5 2 1 4 6 3 15 m3/m4/m5 — — 8 6 5 3 2 16 m1/m3 m3 — 8 6 6 7 2 17 m3 — — 8 4 6 7 1 18 m4/m5 m2/m4 m5 4 3 5 2 6 19 m1/m2/m4/m5 — — 3 6 2 1 3 20 m1/m4 m3/m5 m2/m4/m5 4 7 5 7 7
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表 9 各模具产品的序相关设置时间表
Table 9 The schedule of sequence setup time between each mold
模具 序相关设置时间 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 — 2 3 3 4 3 3 3 3 2 1 3 1 4 4 4 2 4 2 1 2 2 — 2 2 3 2 2 1 3 1 2 2 4 4 3 3 4 4 2 4 3 2 3 — 4 3 3 4 2 1 2 3 3 1 4 2 1 3 3 4 3 4 3 1 3 — 4 1 4 2 2 2 1 1 4 1 3 2 2 1 2 3 5 4 4 4 2 — 1 1 1 4 1 1 1 3 2 1 3 2 4 3 3 6 1 2 3 2 3 — 3 2 2 3 4 1 1 3 3 2 4 2 4 1 7 3 4 4 4 1 2 — 1 1 4 2 2 3 2 2 2 4 1 1 2 8 4 2 4 1 4 1 2 — 4 3 3 3 2 4 1 3 4 1 4 1 9 4 2 3 3 2 4 4 1 — 2 2 3 2 1 2 4 2 1 4 3 10 2 4 1 2 2 4 4 3 3 — 1 1 2 3 1 2 1 4 2 1 11 3 4 3 2 2 2 1 2 2 2 — 2 4 3 3 2 1 2 2 2 12 2 1 4 2 4 2 3 2 1 2 3 — 1 1 3 2 3 2 1 2 13 2 1 2 4 2 1 4 2 4 2 4 3 — 4 4 3 4 3 4 1 14 4 4 1 3 2 2 2 4 1 4 1 2 1 — 2 2 1 3 4 1 15 3 2 2 3 1 2 3 4 3 1 2 1 3 1 — 2 3 1 4 4 16 1 4 1 1 4 3 1 3 1 2 4 4 2 3 4 — 4 4 4 1 17 4 3 1 1 3 2 3 1 3 4 1 1 3 2 1 1 — 2 1 1 18 1 2 4 4 2 1 4 1 2 1 4 4 3 2 1 3 1 — 2 2 19 2 3 2 2 2 3 4 1 1 3 2 2 2 3 3 2 4 1 — 2 20 2 2 2 2 2 2 2 1 2 1 2 4 1 4 1 1 1 4 1 —
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表 10 模具各工序的加工时间表
Table 10 The processing time table of each mold
模具 操作1 操作2 操作3 m1 m2 m3 m4 m5 m1 m2 m3 m4 m5 m1 m2 m3 m4 m5 1 27 — 18 — 29 — — 27 17 — — 23 29 19 — 2 13 — 26 — 11 13 — — — — — — — — — 3 — — 23 — — — — — — — — — — — — 4 — 10 13 — 12 18 25 11 10 26 — 21 31 27 — 5 9 — 8 16 — — — — — — — — — — — 6 — 9 — — 29 19 — — 25 10 30 — — 26 19 7 — 19 — 22 28 — 10 28 24 27 — — — — — 8 30 23 28 — 17 — — — — — — — — — — 9 — — — 31 24 — — 9 — — — — 8 28 — 10 — 25 — — — — 27 — 13 — 31 20 26 — — 11 25 — — — 24 12 — — — — — — — — — 12 — — — 23 — 17 13 16 — — 22 8 — 30 — 13 — 16 — 19 22 — — 28 — — — — — 27 23 14 28 — 16 — 20 — — 18 — — 16 — 16 — 21 15 — — 28 29 23 — — — — — — — — — — 16 27 — 10 — — — — 8 — — — — — — — 17 — — 26 — — — — — — — — — — — — 18 — — — 12 24 — 13 — 27 — — — — — 17 19 24 19 — 12 14 — — — — — — — — — — 20 17 — — 10 — — — 31 — 20 — 30 — 9 24
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表 11 实例仿真结果
Table 11 Simulation results of the instance
运行次数 DTLBO-Ⅱ GA DPSO CCIWO HDTLBO 1 167 166 166 163 163 2 164 166 164 163 166 3 168 167 167 167 167 4 169 166 164 166 163 5 163 163 165 163 165 6 167 165 166 167 163 7 167 168 163 168 166 8 166 168 167 164 168 9 168 167 164 163 163 10 169 168 165 164 165 11 168 163 164 168 166 12 169 168 169 167 163 13 168 167 165 166 167 14 163 170 166 170 166 15 167 166 168 169 163 16 166 167 166 171 163 17 167 170 163 168 163 18 169 168 164 170 165 19 165 168 163 165 168 20 167 168 164 167 163 Average 166.85 166.95 165.15 166.45 164.8
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